
A container $1mole$ of a gas at a temperature ${27^ \circ }C$ has a movable piston which maintains at constant pressure in container of $1atm$ . The gas is compressed until temperature becomes ${127^ \circ }C$ . The work done is?
(C for gas is $7.03cal/molK$ )
A. $703J$
B. $814J$
C. $121J$
D. $2035J$
Answer
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Hint: Since the gas is compressed at a constant pressure to reach a specific temperature in the given problem and work done varies with the conditions of the system given, therefore, the work done can be calculated by using the ideal gas equation and pressure-volume work during the process of temperature change.
Formula used:
The formulae used in this problem for the calculation of work done is: -
$Workdone = W = P\Delta V$
$P\Delta V = nR\Delta T$
Complete answer:
Since the gas in a container with movable piston is compressed at constant atmospheric pressure of $P = 1atm$ (given)
Initial Temperature of gas, ${T_1} = {27^ \circ }C = 300K$ $\left( {^ \circ C + 273 = K} \right)$
Gas is compressed until it reaches the final temperature of ${T_2} = {127^ \circ }C = 400K$ (given)
Therefore, change in temperature will be $\Delta T = {T_2} - {T_1} = 400 - 300 = 100K$
Now, we know that Pressure-Volume Work in thermodynamics is defined as: -
At constant pressure, $Workdone = W = P\Delta V$ … (1)
But Ideal-Gas Equation is given as: $P\Delta V = nR\Delta T$ … (2)
Equating (1) and (2), work done will be: -
$W = nR\Delta T$ … (3)
where,
$R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}$ = Gas Constant
$n = 1$ = no. of moles (given)
Substituting these values in eq. (3), we get
$W = 1 \times 8.314 \times \Delta T$
Substituting the value of $\Delta T$ in the above expression, we get
$ \Rightarrow W = 1 \times 8.314 \times 100$
$ \Rightarrow W = 831.4J$
Approximating the values according to the options given, we get
$ \Rightarrow W \approx 814J$
Thus, the work done in compressing the gas is about $814J$ approximately. Hence, the correct option is (B) $814J$.
Note: In this problem, to calculate the amount of work done in compressing a gas using a moving piston, it is necessary to create a relationship between work done and temperature change using the ideal gas equation and pressure-volume work. Always remember to study the provided parameters while handling this type of numerical problem in order to provide a precise answer.
Formula used:
The formulae used in this problem for the calculation of work done is: -
$Workdone = W = P\Delta V$
$P\Delta V = nR\Delta T$
Complete answer:
Since the gas in a container with movable piston is compressed at constant atmospheric pressure of $P = 1atm$ (given)
Initial Temperature of gas, ${T_1} = {27^ \circ }C = 300K$ $\left( {^ \circ C + 273 = K} \right)$
Gas is compressed until it reaches the final temperature of ${T_2} = {127^ \circ }C = 400K$ (given)
Therefore, change in temperature will be $\Delta T = {T_2} - {T_1} = 400 - 300 = 100K$
Now, we know that Pressure-Volume Work in thermodynamics is defined as: -
At constant pressure, $Workdone = W = P\Delta V$ … (1)
But Ideal-Gas Equation is given as: $P\Delta V = nR\Delta T$ … (2)
Equating (1) and (2), work done will be: -
$W = nR\Delta T$ … (3)
where,
$R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}$ = Gas Constant
$n = 1$ = no. of moles (given)
Substituting these values in eq. (3), we get
$W = 1 \times 8.314 \times \Delta T$
Substituting the value of $\Delta T$ in the above expression, we get
$ \Rightarrow W = 1 \times 8.314 \times 100$
$ \Rightarrow W = 831.4J$
Approximating the values according to the options given, we get
$ \Rightarrow W \approx 814J$
Thus, the work done in compressing the gas is about $814J$ approximately. Hence, the correct option is (B) $814J$.
Note: In this problem, to calculate the amount of work done in compressing a gas using a moving piston, it is necessary to create a relationship between work done and temperature change using the ideal gas equation and pressure-volume work. Always remember to study the provided parameters while handling this type of numerical problem in order to provide a precise answer.
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